Question

The half-life of sodium-24 is 15.0 hours. What percentage of a sample of sodium-24 remains after 75.0 hours?

Answer 1

3.124 %

Step-by-step explanation:

Given that:

Half life = 15.0 hours

`t_(1/2)=\frac {ln\ 2}{k}`

Where, k is rate constant

So,

`k=\frac {ln\ 2}{t_(1/2)}`

`k=\frac {ln\ 2}{15.0}\ hour^(-1)`

The rate constant, k = 0.04621 hour⁻¹

Time = 75.0

Using integrated rate law for first order kinetics as:

`[A_t]=[A_0]e^(-kt)`

Where,

`[A_t]` is the concentration at time t

`[A_0]` is the initial concentration

So,

`\frac {[A_t]}{[A_0]}=e^(-0.04621* 75)`

`\frac {[A_t]}{[A_0]}=0.03124`

In percentage, the sample of sodium-24 remains = 3.124 %